A train passes through a station with a constant speed. A stationary observer at the station platform measures the tone of the train whistle as 484 Hz when it approaches the station and  442 Hz when it leaves the station. If the sound velocity in air is  330m/s, then the tone of the whistle and the speed of the train are

When train approaches the station, the frequency heard by the observerυ1=υvv−vs=υ330330−vsHere,  v=330 m/sυ is the actual frequency of the whistle484=υ330330−vs.....1When the train leaves the stationυ2=υvv+vs=υ330330+vs....2Divide Equations (1) & (2), we get⇒484442=330+vs330−vs⇒1.09=330+vs330−vs⇒vs=31.352.09=15 m/s=15 ×185=54kmphSubstituting  vs in equation gives⇒484=υ330330−15=υ330315⇒υ=484×2122∴υ=462 HzTherefore, the correct answer is (A).